Question

In series with B.G. of resistance $30\,\Omega$. The coil is placed with its plane perpendicular to the direction of a uniform magnetic field of induction $10^{-2}$ T. If it is now turned through an angle of $60^{\circ} $ about an axis in its plane. Find the charge induced in the coil. (Area of a coil $=10^{-2} m^2)$

• A. $2\times10^{-5} C$
• B. $3.2\times10^{-5} C$
• C. $1\times10^{-5} C$
• D. $5.5\times10^{-5} C$

Answer 1

Answer: (C)

Given : $n =10$ turns, $R_{\text {coil }}=20\, \Omega, R_{G}=30 \,\Omega$,
Total resistance in the circuit $=20+30$
$=50 \,\Omega$
$A=10^{-2} m ^{2}, B=10^{-2} T$,
$ \phi=0^{\circ}, \phi_{2}=60^{\circ}$
$q=\frac{\phi_{1}-\phi_{2}}{R}=\frac{B n A \cos \theta_{1}-B n A \cos \theta_{2}}{R}$
$=\frac{B n A(\cos 0-\cos 60)}{R}=\frac{B n A(1-0.5)}{R}$
$= \frac{0.5 \times 10^{-2} \times 10 \times 10^{-2}}{50}$
$=\frac{50 \times 10^{-5}}{50}$
$1 \times 10^{-5} C$